1. Find the HCF of a) 21 and 8 $$ \begin{array}{l} 21- \ 8- \end{array} $$ b) 108 and 270 $$ \begin{array}{l} 108- \ 270- \end{array} $$ 416 and 8 6- 8- d) $$ 12-9 $$ $$ 12- $$ $$ 4- $$

Question

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Sure, let's find the Highest Common Factor (HCF) of each pair of numbers using the **Euclidean Algorithm**. The Euclidean Algorithm is an efficient method for computing the HCF of two numbers based on the principle that the HCF of two numbers also divides their difference.

### a) HCF of 21 and 8

**Step-by-Step Process:**

1. **Divide 21 by 8:**
   $$
   21 \div 8 = 2 \quad \text{with a remainder of} \quad 21 - (8 \times 2) = 5
   $$
   $$
   21 = 8 \times 2 + 5
   $$

2. **Now, divide 8 by the remainder 5:**
   $$
   8 \div 5 = 1 \quad \text{with a remainder of} \quad 8 - (5 \times 1) = 3
   $$
   $$
   8 = 5 \times 1 + 3
   $$

3. **Next, divide 5 by the remainder 3:**
   $$
   5 \div 3 = 1 \quad \text{with a remainder of} \quad 5 - (3 \times 1) = 2
   $$
   $$
   5 = 3 \times 1 + 2
   $$

4. **Then, divide 3 by the remainder 2:**
   $$
   3 \div 2 = 1 \quad \text{with a remainder of} \quad 3 - (2 \times 1) = 1
   $$
   $$
   3 = 2 \times 1 + 1
   $$

5. **Finally, divide 2 by the remainder 1:**
   $$
   2 \div 1 = 2 \quad \text{with a remainder of} \quad 2 - (1 \times 2) = 0
   $$
   $$
   2 = 1 \times 2 + 0
   $$

**Conclusion:**
When the remainder becomes **0**, the last non-zero remainder is the HCF.
$$
\boxed{1}
$$

---

### b) HCF of 108 and 270

**Step-by-Step Process:**

1. **Divide 270 by 108:**
   $$
   270 \div 108 = 2 \quad \text{with a remainder of} \quad 270 - (108 \times 2) = 54
   $$
   $$
   270 = 108 \times 2 + 54
   $$

2. **Now, divide 108 by the remainder 54:**
   $$
   108 \div 54 = 2 \quad \text{with a remainder of} \quad 108 - (54 \times 2) = 0
   $$
   $$
   108 = 54 \times 2 + 0
   $$

**Conclusion:**
When the remainder becomes **0**, the last non-zero remainder is the HCF.
$$
\boxed{54}
$$

---

### c) HCF of 416 and 8

**Step-by-Step Process:**

1. **Divide 416 by 8:**
   $$
   416 \div 8 = 52 \quad \text{with a remainder of} \quad 416 - (8 \times 52) = 0
   $$
   $$
   416 = 8 \times 52 + 0
   $$

**Conclusion:**
Since the remainder is already **0**, the divisor at this step is the HCF.
$$
\boxed{8}
$$

---

### d) HCF of 12 and 9

**Step-by-Step Process:**

1. **Divide 12 by 9:**
   $$
   12 \div 9 = 1 \quad \text{with a remainder of} \quad 12 - (9 \times 1) = 3
   $$
   $$
   12 = 9 \times 1 + 3
   $$

2. **Now, divide 9 by the remainder 3:**
   $$
   9 \div 3 = 3 \quad \text{with a remainder of} \quad 9 - (3 \times 3) = 0
   $$
   $$
   9 = 3 \times 3 + 0
   $$

**Conclusion:**
When the remainder becomes **0**, the last non-zero remainder is the HCF.
$$
\boxed{3}
$$

---

**Summary of HCFs:**

- **a) HCF of 21 and 8:** 1
- **b) HCF of 108 and 270:** 54
- **c) HCF of 416 and 8:** 8
- **d) HCF of 12 and 9:** 3
- a) HCF of 21 and 8: 1 - b) HCF of 108 and 270: 54 - c) HCF of 416 and 8: 8 - d) HCF of 12 and 9: 3

Find the HCF of
a) 21 and 8

b) 108 and 270

416 and 8
6-
8-
d)

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Find the HCF of a) 21 and 8 b) 108 and 270 416 and 8 6- 8- d)

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Latest Arithmetic Questions

Find the HCF of
a) 21 and 8

b) 108 and 270

416 and 8
6-
8-
d)